Tìm x thuộc N biết
a) (x+3)2=16
Tìm x thuộc N biết
a) \(\left(2x-1\right)^3=4^{12}:16^5\)
b) 6x + 5 chia hết cho (3x - 1)
a: =>(2x-1)^3=4^12:4^10=4^2=8
=>2x-1=2
=>2x=3
=>x=3/2(loại)
b: 6x+5 chia hết cho 3x-1
=>6x-2+7 chia hết cho 3x-1
=>7 chia hết cho 3x-1
mà x là số tự nhiên
nên 3n-1=-1
=>n=0
tìm x thuộc n biết
A. 3\(^x\)=81x3
b.2\(^{x+1}\)=32
c. 3\(^{x+2}\):27=3
d. 2x2=32
e. (2x-1)4=81
f. (2x-6)4=0
a: =>3^x=3^4*3=3^5
=>x=5
b: =>\(2^{x+1}=2^5\)
=>x+1=5
=>x=4
c: \(\Leftrightarrow3^{x+2-3}=3\)
=>x-1=1
=>x=2
d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)
=>x=4 hoặc x=-4
e: (2x-1)^4=81
=>2x-1=3 hoặc 2x-1=-3
=>2x=4 hoặc 2x=-2
=>x=-1 hoặc x=2
f: (2x-6)^4=0
=>2x-6=0
=>x-3=0
=>x=3
a) \(3^x=81\cdot3\)
\(\Rightarrow3^x=3^4\cdot3\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
b) \(2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
c) \(3^{x+2}:27=3\)
\(\Rightarrow3^{x+2}:3^3=3\)
\(\Rightarrow3^{x+2-3}=3\)
\(\Rightarrow3^{x-1}=3\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
d) \(2x^2=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
e) \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
f) \(\left(2x-6\right)^4=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=6:2\)
\(\Rightarrow x=3\)
\(a,3^x=81\cdot3\\ \Leftrightarrow3^x=3^4\cdot3\\ \Leftrightarrow3^x=3^5\\ \Leftrightarrow x=5\\ d,2^{x+1}=32\\ \Leftrightarrow x+1=5\\ \Leftrightarrow x=4\\ c,3^{x+2}:27=3\\ \Leftrightarrow3^{x+2}:3^3=3\\ \Leftrightarrow3^{x-1}=3\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,2x^2=32\\ \Leftrightarrow x^2=16\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ e,\left(2x-1\right)^4=81\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ f,\left(2x-6\right)^4=0\\ \Leftrightarrow2x-6=0\\ \Leftrightarrow x=3\)
tìm x , biết
a) 17/6- x( x-7/6)= 7/4
b) 3/35 - ( 3/5-x)= 2/7
tìm x thuộc Z , biết
3/4-5/6 < x/12 < 1 -( 2/3-1/4)
tìm x biết
a ) 2x-3=x + 1/2
b) 4x- ( x+ 1/2) = 2x - ( 1/2 - 5 )
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
Bài 3:
a) Ta có: \(2x-3=x+\dfrac{1}{2}\)
\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)
\(\Leftrightarrow x=5\)
Bài 1: Tìm x thuộc N, biết
a) x=x mũ 5
b)x mũ 4= x mũ 2
c)(x-1)mũ 3 = x-1
Bài 2: Tìm x
(2x -1) mũ 3= 1 mũ 3+ 2 mũ 3+3 mũ 3+ 4 mũ 3+ 5 mũ 3
Bài 1
a) \(x=x^5\)
\(x^5-x=0\)
\(x\left(x^4-1\right)=0\)
\(x=0\) hoặc \(x^4-1=0\)
* \(x^4-1=0\)
\(x^4=1\)
\(x=1\)
Vậy x = 0; x = 1
b) \(x^4=x^2\)
\(x^4-x^2=0\)
\(x^2\left(x^2-1\right)=0\)
\(x^2=0\) hoặc \(x^2-1=0\)
*) \(x^2=0\)
\(x=0\)
*) \(x^2-1=0\)
\(x^2=1\)
\(x=1\)
Vậy \(x=0\); \(x=1\)
c) \(\left(x-1\right)^3=x-1\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)
*) \(x-1=0\)
\(x=1\)
*) \(\left(x-1\right)^2-1=0\)
\(\left(x-1\right)^2=1\)
\(x-1=1\) hoặc \(x-1=-1\)
**) \(x-1=1\)
\(x=2\)
**) \(x-1=-1\)
\(x=0\)
Vậy \(x=0\); \(x=1\); \(x=2\)
tìm x biết
a) (2x-3)(2x+3)=0
b) x^2-1=0
c) x^2-9=0
d) 4^2-16=0
e) 25x^2-9=0
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Tìm x thuộc N ,biết
a ,11 chia hết cho ( x + 1 )
b , x + 8 chia hết cho x +1
c , x + 11 chia hết cho x + 2
a: \(x+1\in\left\{1;11\right\}\)
hay \(x\in\left\{0;10\right\}\)
b: \(\Leftrightarrow x+1\in\left\{1;7\right\}\)
hay \(x\in\left\{0;6\right\}\)
Tìm số tự nhiên x biết
a,3 . x - 15 = x + 35
b, (8.x-16).(x-5)=0
c, x.(x+1)= 2 + 4 + 6 + 8 +10+......+2500
\(a,3\cdot x-15=x+35\)
\(\Rightarrow3x-x=35+15\)
\(\Rightarrow 2x=50\)
\(\Rightarrow x = 50:2\)
\(\Rightarrow x= 25\)
\(b,(8x-16)(x-5)=0\)
\(+, TH1: 8x-16=0\)
\(\Rightarrow8x=16\)
\(\Rightarrow x = 16:8\)
\(\Rightarrow x=2\)
\(+,TH2: x-5=0\)
\(\Rightarrow x =5\)
\(c,x(x+1)=2+4+6+8+10+...+2500\) \(^{\left(1\right)}\)
Đặt \(A=2+4+6+8+10+...+2500\)
Số các số hạng của \(A\) là: \(\left(2500-2\right):2+1=1250\left(số\right)\)
Tổng \(A\) bằng: \(\left(2500+2\right)\cdot1250:2=1563750\)
Thay \(A=1563750\) vào \(^{\left(1\right)}\), ta được:
\(x\left(x+1\right)=1563750\)
\(\Rightarrow x\left(x+1\right)=1250\cdot1251\)
\(\Rightarrow x =1250\)
#\(Toru\)
tìm x,y thuộc Z biết
a,(3x+2).(y-8)=12
b,(5x-4).(y+3)=-18
\(a,\) Vì \(x,y\in Z\) nên \(\left(3x+2\right):3R2;R1\)
Mà \(\left(3x+2\right)\left(y-8\right)=12\) nên \(3x+2\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Do đó \(3x+2\in\left\{-4;-1;2\right\}\)
\(\Rightarrow x\in\left\{-2;-1;0\right\}\)
Với \(x=-2\Rightarrow\left(-4\right)\left(y-8\right)=12\Rightarrow y-8=-3\Rightarrow y=5\)
Với \(x=-1\Rightarrow\left(-3\right)\left(y-8\right)=12\Rightarrow y-8=-4\Rightarrow y=4\)
Với \(x=0\Rightarrow2\left(y-8\right)=12\Rightarrow y-8=6\Rightarrow y=14\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-2;5\right);\left(-1;4\right);\left(0;14\right)\)
\(b,\) Vì \(x,y\in Z\) nên \(\left(5x-4\right):5R1;R4\)
Mà \(\left(5x-4\right)\left(y+3\right)=-18\)
\(\Rightarrow5x-4\inƯ\left(-18\right)=\left\{-18;-9;-6;-3;-2;-1;1;2;3;6;9;18\right\}\\ \Rightarrow5x-4\in\left\{-9;1;6\right\}\\ \Rightarrow x\in\left\{-1;1;2\right\}\)
Với \(x=-1\Rightarrow-9\left(y+3\right)=-18\Rightarrow y+3=2\Rightarrow y=-1\)
Với \(x=1\Rightarrow y+3=18\Rightarrow y=15\)
Với \(x=2\Rightarrow6\left(y+3\right)=18\Rightarrow y+3=3\Rightarrow y=0\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-1;-1\right);\left(1;15\right);\left(2;0\right)\)
Tìm x,y ∈ N , biết
a.( x + 2 ) . ( y + 3 ) = 6
b.( x - 3 ) . ( y + 1 ) = 7
\(a,\text{Vì }x,y\in N\Leftrightarrow x+2\ge2;y+3\ge3\\ \Leftrightarrow\left(x+2\right)\left(y+3\right)=6=2\cdot3=3\cdot2\\ \Leftrightarrow\left\{{}\begin{matrix}x+2=2\\y+3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(0;0\right)\)
\(b,\Leftrightarrow\left(x-3\right)\left(y+1\right)=7\cdot1=1\cdot7\\ \left\{{}\begin{matrix}x-3=7\\y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=0\end{matrix}\right.\\ \left\{{}\begin{matrix}x-3=1\\y+1=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(10;0\right);\left(4;6\right)\right\}\)
Câu 2: Tìm x, biết
a) x + 72 = 36
b) 16 x2 = 64
c) (5.x – 2) – 64 = -36
d) (2x-10) . (5 – x) = 0
giúp mình với mọi người
a: x=36-72=-36
d: =>x-5=0
hay x=5
a) x + 72 = 36
x = 36 - 72 = -36
b) 16 x2 = 64
x2 = 64 : 16 = 4
x2 = 22
→ x = 2
c) (5 . x - 2) - 64 = -36
(5 . x - 2) = -36 + 64
5 . x - 2 = 28
5 . x = 28 + 2 = 30
x = 30 : 5 = 6
d) (2x - 10) . (5 - x) = 0
x = 5